Two capillary tubes of same length each of $50 \ cm$ but of different radii $4 \ mm$ and $2 \ mm$ are connected in series. When water flows,the pressure difference between the ends of the arrangement is $P$. Then the pressure difference between the ends of the first tube is

Two capillary tubes of the same radius but lengths $l_1$ and $l_2$ are connected in parallel at the bottom of a vessel. What should be the length of a single capillary tube of the same radius that replaces both,such that the flow rate remains the same?

In Poiseuille's method for the determination of the coefficient of viscosity,the physical quantity that requires the greatest accuracy in measurement is:

Consider a steady flow of oil in a pipeline. The cross-sectional radius of the pipeline decreases gradually as $r = r_0 e^{-\alpha x}$, where $\alpha = \frac{1}{3} \text{ m}^{-1}$ and $x$ is the distance from the pipeline inlet. If $R_1$ is the Reynolds number for a certain pipeline cross-section at a distance $x_1$ metre from the inlet and $R_2$ is for distance $(x_1 + 3)$ metre, then the ratio $\frac{R_1}{R_2}$ is

In the case of a liquid, if the Reynolds number '$R_n$' is $1450$, then the flow of the liquid will

For which of the following Reynolds numbers is a flow considered streamlined (laminar)?